text:
May aerobatics be performed over a controlled airport without getting a waiver?
CFR 14 Part 91.303 states:

Aerobatic flight.
No person may operate an aircraft in aerobatic flight--
(a) Over any congested area of a city, town, or settlement;
(b) Over an open air assembly of persons;
(c) Within the lateral boundaries of the surface areas of Class B, Class C, Class D, or Class E airspace designated for an airport;
(d) Within 4 nautical miles of the center line of any Federal airway;
(e) Below an altitude of 1,500 feet above the surface; or.
(f) When flight visibility is less than 3 statute miles.
For the purposes of this section, aerobatic flight means an intentional maneuver involving an abrupt change in an aircraft's attitude, an abnormal attitude, or abnormal acceleration, not necessary for normal flight.

For the purposes of this question I want to focus on paragraph (c). Assuming we don't have any other specific permission (waiver), may aerobatics be performed above class B, C, D, or E surface airspace?
I know aerobatics may be performed under the outer rings of class B and C airspace.
There also doesn't appear to be anything preventing aerobatics above the outer rings of class B or C airspace.
So the question is focusing on the center core (the surface area) of class B and C as well as the typical cylinder of a class D airspace and whatever shape a class E surface area happens to be.
Paragraph (c) simply states "Within the lateral boundaries of the surface areas" but makes no mention of altitude (other than the 1500' AGL reference in paragraph (e)).
Obviously we are talking about surface areas that go to the ground so clearly you may not fly in or below these surface areas. But may you fly above them as long as you are 1500' AGL?
"within the lateral boundaries of" has no implied altitude restriction. That term, in English, implies surface to outer space. So it would seem that aerobatics may not be performed above a class B, C, D, or E surface area at any altitude.
But you may perform aerobatics above the outer rings of class B and C. So why restrict doing them over the surface area?
Is there an official answer? May I legally do aerobatics above class B, C, D, or E surface airspace?
I have searched and looked at many related topics but none address this specific question. Even the Hucker discussion leaves this unanswered (to me at least) which only clarifies being able to do aerobatics under (no mention of over) the outer rings of class B and C airspace.

You can with a waiver, even on the airfield itself. A lot of airshows operate this way.

@RonBeyer I understand that. I specifically stated at the start of my question that this is assuming no other specific permission.

To me it seems like you answered your own question: "within the lateral boundaries of" has no implied altitude restriction. That term, in English, implies surface to outer space. So it would seem that aerobatics can't be performed above a class B, C, D, or E surface area at any altitude.

Sometimes the FARs are ambiguous, but in this case it’s clear. Since, lateral is not defined in 14 CFR §1, the AIM, or the Pilot/Controller Glossary you revert to the dictionary definition. The rule means exactly what it says. You can’t do acrobatics within the lateral boundaries of the airspace.

@J.Hougaard There's a difference between what I think it means and what it really means. I'm looking for a definitive, legal answer, not just what people think it means. If someone climbs over a class D airspace, for example, will they get busted if they do aerobatics there?

If you want a legal answer, you'll have to get that in an opinion letter from the FAA itself. I don't think there is any precedent set (rule clarifications). This usually starts with contacting your local FSDO directly. Unfortunately we can't give legal advise, nor would it stand up in a court.

@RonBeyer True. So for this question I'll settle for precedent. Does anyone know of people that have regularly conducted aerobatics above a controlled airport without prior permission and no risk of being busted?

@rmaddy Citing examples of people that have done X or Y is not usually a good way of determining how to interpret the FARs. I could give you examples of old crop dusters doing loops and rolls in a C177 without a parachute or much regard for the prohibitions of §91.303. And the risk of a pilot getting busted generally depends on the legality of what is being done, which depends on what the regulations say and how they are interpreted. I think what you need is a legal interpretation, not examples of what has or has not been done.

@JonathanWalters I knew that statement would get some pushback. :) What I meant are examples where a group of pilots have been doing aerobatics above airports for years, with full knowledge of relevant authorities, and without any special permission beyond what is in the FARs. I am certainly not looking for one-off examples where a pilot "got away with it".

Can they, sure....you are completely capable of doing so......should you though????

@NZKshatriya I'm only looking for the "can" part of the answer, not the "should". On what basis do you state that aerobatics can (legally) be performed above the lateral boundaries of the surface areas of Class B, Class C, Class D, or Class E airspace designated for an airport without any other specific permission?

@rmaddy  sigh  I was actually commenting on the improper usage of can in this situation.  Can refers to one's ability to do something, while may refers to one's permission or right to do something.

@NZKshatriya OK, point taken. Your use of "should" in your first comment through me off the point you were trying to make. I'll update the question to properly use "may" where I mistakenly used "can".

@rmaddy  I had just finished time keeping a speech competition.  And let's just say that grammar and English usage skills were not up to par (this was a college level competition for poetry).  I was a little bit miffed, sorry for getting on your, I sort of vented in the wrong place lol.

Actually "within the lateral boundaries of the surface areas of any Class D, Class C, or Class B airpace" DOES imply that you are not permitted to do aerobatics over these areas without special permits from the FAA or if there is a designated aerobatic box at that airport and said pilot remains within that airspace.  Even if you are not in the actual airspace itself, if you're above it, you're not allowed to fly acro over it.
Now, note the term "surface areas" could imply that you could fly aerobatics above or below Class C or Class B airspace SHELVES, that is a section of Class C or Class B airspace which does not touch the ground,  provided you don't enter those areas and said airspace meets all the other requirements of FAR 91.303.

See Hucker (2006) which states unequivocally that flight above Class B and C shelves is not prohibited.

Right. I agree with the implication. That's what I'm trying to answer clearly. Do people practice aerobatics in said airspace without any special permission beyond what's in the FAR and do so without risk of being busted?

@Jonathan I assume you are clarifying this answer because I covered that in my question. And my question is not about flying over the shelves since that aspect is clear. Let's not spend time on that. Thanks.

@rmaddy Yes. This answer indicates that the author did not fully read your question and is not familiar with Hucker (2006).

Can someone provide a working link to Hucker (2006)?

@quietflyer I cant’t find it on the FAA site anymore. This is my copy. https://www.touringmachine.com/LegalInterpretations/Legal%20Interpretation%20Hucker%20-%20Aerobatics%20Class%20B(2006).pdf

Reasons for rollback: 1) when I made the edit I was new to ASE and unfamiliar w/ the culture; this is way too much content to add to someone else's post.  2) I no longer agree with statements like "At any rate, the Class-E-to-surface "extensions" do not fall within the meaning of "Within the lateral boundaries of the surface areas of... Class E airspace designated for an airport"."  The matter is ambiguous.  For more see related answer https://aviation.stackexchange.com/a/55806/34686 .

Re "provided you don't enter those areas" (last sentence) -- why could you not enter the shelves?

@JW -- re "See Hucker (2006) which states unequivocally that flight above Class B and C shelves is not prohibited." -- actually, it doesn't.  I just re-read it and the issue of flight above the ceiling of Class B, C, or for that matter D is actually not directly addressed.

(Obviously answerer is free to re-add rolled-back content if wishes-- I just don't want it attributed to me! )  ; )

Consider the portion of class B, C, D, or E airspace that actually extends all the way to the ground.  (In the case of Class E airspace, it is arguably ambiguous as to whether or not surface-level E3/E4 "extensions" that do not actually surround the airport in question qualify as "Class E airspace designated for an airport".1)
The plain English meaning of "within the lateral boundaries of" does not imply any vertical limit of any kind.
Also, the FAA uses the term "surface area" to mean not just an area on a map, but rather a volume of airspace whose floor is actually in contact with the surface.3
So for all the portions of class B, C, D, or E airspace that actually extend all the way to the ground (again recognizing the Class E "extensions" fall into an ambiguous category), there is a cylinder (or more irregularly shaped column) of airspace going all the way to the top of the atmosphere and beyond -- or at least to the limit of the FAA's jurisdiction-- within which aerobatics may not be performed without a waiver, by virtue of FAR 91.303(c).
Aerobatics over or under or even within4 class B or C "shelf" airspace are fine, as long as you are not within the cylinder or column described above, extending upward indefinitely from controlled airspace that is actually in contact with the surface.
For more, see the 2006 Hucker interpretation, which reversed an earlier 1999 interpretation.5  The truth is, however, that the 2006 Hucker interpretation does not specifically address the issue of doing aerobatics above the ceiling of Class B, C, or D airspace.  Instead we have to rely on the common-sense meaning of "within the lateral boundaries of...".

But you may perform aerobatics above the outer rings of class B and C.
So why restrict doing them over the surface area?

That is a very good question.  The ways of the FAA are unfathomable to mere mortals.
Footnotes:

I'll refrain from getting deeper into the controversial issue of surface-level Class E "extensions" (E3/E4 airspaces) here.  To explore more, readers may wish to start with the related ASE answer "Does FAR 91.155c apply to class E surface extensions?", and then check out the additional links given at the end of that answer.

For a list of 11 other FARs whose wording contains the "within the lateral boundaries of" phrasing, see the related ASE question What indication has the FAA given that phrases like "surface area of Class E airspace designated for an airport" do or don't include E4 "extensions"?. Even if the reader has no interest in the "extension" issue, he or she may find the list of FARs to be of interest.  Note that in some of these FARs, the "within the lateral boundaries of..." phrasing is only applied to surface-level Class E airspace, while in others of these FARs it is applied to surface-level controlled airspace in general.  Two more FARs that use the "within the lateral boundaries of..." language, but apply it to the entirety of Class B and Class C airspace (which would include "shelves"), with no reference to surface area, are FAR 91.215(b)(4) and  FAR 91.225(d)(3).

Re the FAA's use the term "surface area", sometimes capitalized as "Surface Area" -- see for example the language of the "Airspace Designations and Reporting Points" document quoted at the bottom of Page 1 of the FAA Memorandum reproduced in the ASE answer linked in footnote 1.  See also the reference to ATC/FSS use of the phrase "surface area" in the text immediately above the map figure in that answer.  Clearly in these cases we are talking about a volume of airspace whose floor is in contact with the ground, not just an area on the ground.  However in reference to the present question, we'd come to the same conclusion even if the phrase were simply being used to describe an area on the ground, due the "within the lateral boundaries" phrasing.  The location of the ceiling of the airspace called the "surface area of controlled airspace", or the question of whether "surface area" means a volume of airspace or an area on the ground, is actually irrelevant to the interpretation of FAR 91.303(c) and other FARs containing similar "within the lateral boundaries of" language. This point pertains especially to some of the answers given to the ASE question Which parts of class E airspace can an ultralight (part 103) fly in without prior ATC authorization? -- at least one answer really "misses the boat" on this point.  (In fact, on at least one occasion, the FAA itself "missed the boat" on this point-- an October 1993 Final Rule pertaining to FAR 91.157(a) implied that the phrase "within the lateral boundaries of" somehow also implied a vertical limit, and therefore changed the language of FAR 91.157(a) to read "...within the airspace contained by the upward extension of the lateral boundaries of the controlled airspace designated to the surface...". It's odd that they didn't see a need to change the similar language in 91.303(a), as well as all the other FARs with similar language noted in footnote 2 of this answer.  For details see this ASE answer to the ASE question "Which parts of class E airspace can an ultralight (part 103) fly in without prior ATC authorization?".)

Obviously, if you are within Class B or C airspace, you are subject to other constraints, namely instructions from ATC, which may prevent you from doing aerobatics without a waiver, but that has nothing to do with FAR 91.303(c).  Note that the text of the Hucker ruling specifically states that the questioner asked "whether aerobatics may be performed within Class B airspace (with proper ATC clearance), and/or underneath the floors of Class B airspace."  The ruling stated "aerobatics may be conducted in such outer areas (i.e. within the "shelves" that do not extend to the ground), provided that the requirements of 91.130 or 91.131 are met as applicable; and the requirements of 91.303 (a), (b), (d), (e) and (f) are met."

Apparently the text of this older 1999 decision is not currently available on the FAA's website.  A search for Legal Interpretations using the key word "aerobatics" only yields results going back to the year 2000.

Answer needed to be updated to fix link to "Hucker" interpretation, and I took the opportunity to make some other changes. What this answer adds to the other (highly upvoted) one-- more correctly addresses aerobatics within Class B and C shelves. Also, offers a stronger statement than "Now, note the term "surface areas" could imply that you could fly aerobatics above or below Class C or Class B airspace SHELVES" (itals added)-- you absolutely can, and within the shelves as well.

Actually, I only just now realized as of last (and final?) edit, that the 2006 Hucker interpretation doesn't actually specifically address aerobatics above the ceiling of Class B, C, or D airspace, various comments under other answer notwithstanding.

Prefer Hucker "interpretation" to "ruling" or "decision"-- thought had changed in all instances, but missed at least one-- won't fix right now though. Ditto for older 1999 interpretation.  (See footnotes 4 and 5.)

It's a shame that the inquiries that led to the issuances of Interpretations from the Office of the chief Counsel are no longer viewable-- they used to be. Now one can only see the Interpretations. Anyway, I believe I've found an FAA interpretation that is specific to aerobatics that bears on the "extension" question. If able to confirm, that info deserves to be incorporated into this answer. Stay tuned. (Also, mention "itals added" fn 4.)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the load factor in a descending turn?
Possibly related to How does the load factor vary when the aircraft pitches up/down?
Could you help to settle a debate? I have recently flown a dead-stick approach with a CFI in the right seat. Turning base I was slightly high, and deliberately went past the final track, then banked to around 30 degrees and re-intercepted from the other side. During the turn, the plane was coordinated and I didn't pull up; I've merely altered back-pressure to keep the airspeed stable.
My first CFI taught me that whilst descending I can essentially steep-bank the plane provided I don't pull up to the extent that the airspeed drops (ie not a level turn).
Due to the descent the load factor wouldn't increase, at least not to the extent it does in a level turn.
The new CFI was of a different opinion. Which one is true, or is it somewhere in the middle?
For the setup:
In a level turn, 30 degrees would equate to 1.15G and increase the stall speed by 7%.  What are these values if at the initiation of the turn, the aircraft was in a stable (call it 600fpm) descent, and the airspeed was kept constant (relaxing back pressure to not decrease or increase the speed)?

You might be on the wrong path in discussing load-factor. You may be more interested in stall behavior, which is indirectly related to load-factor but not synonymous. I say this because you are discussing an approach which generally means you are well below Va and thus load factor is not really relevant. What is relevant is the AOA and momentary vertical(relative to earth) acceleration.
As a bit of an aside the most efficient use of availible lift for a turn to a heading is around 45deg bank and best-glide range AOA. Best bank and speed varies some based on the details of real aircraft.

Load Factor (in the sense of wing loading) is the ratio of Lift to Weight.

Assume for simplicity Thrust is zero.  (If Thrust were not zero, we would simply have to substitute the "Excess Drag" vector, defined as (Drag - Thrust), in place of the Drag vector, in all occurrences below.)

Assume "coordinated" flight, meaning zero sideslip, meaning that none of the aircraft's weight is being supported by an aerodynamic sideforce generated by the air striking the side of the fuselage, the side of the vertical fin, etc.

In a steady-state descent, we can draw a closed vector triangle between the Weight vector, the Drag vector, and the non-centripetal component of the Lift vector.

For any given bank angle, the ratio between the total Lift vector and the non-centripetal component of the Lift vector is fixed.

For brevity we'll use the term "Nclift" to mean the non-centripetal component of the Lift vector.

Nclift = Lift * cosine (Bank)

Lift = Nclift / cosine (Bank)

Load Factor = Nclift / (Weight * cosine Bank)

In a steady-state descent, (Nclift)^2 + (Drag)^2 = (Weight)^2.  (This particular expression actually doesn't figure into the math below.)

Also, in a steady-state descent, Nclift / Weight = cosine (arctan (D / Nclift)), so Nclift = cosine (arctan (D / Nclift)) * Weight

The expression that we're taking the cosine of in the line above, is also the glide angle, i.e. the angle of descent.

Which means that in a steady-state descent, Load Factor = Lift/Weight = (Nclift / Weight)/ cosine bank =  (cosine (arctan (D / Nclift))) / cosine bank = (cosine(arctan(D / (L * cosine bank))) / cosine bank

As we degrade the L/D ratio, and thus also the ratio of Nclift / D, we carry some of the aircraft's Weight with Drag vector rather than the Nclift vector.  For any given bank angle, in a steady-state descent, lower values of L/D therefore must be associated with lower values of L, due to a decrease in airspeed, angle-of-attack, or both.  So "dirtying up" the aircraft by dropping flaps, etc -- and therefore making the glide path steeper-- does force a reduction in Lift, and thus a reduction in G-loading.  But how significant is this effect really, for "normal" aircraft in "normal" situations?

Let's assume we can vary the ratio of L/D from infinity, down to 8/1, or down to 4/1, or down to 2/1, by "dirtying up" the airplane.  How do these 4 cases compare, in terms of the load factor we'd have at any given bank angle?

(Remember, an infinite L/D ratio in this context would simply represent a case where we did in fact have some Thrust, exactly equal to Drag, since the variable we're calling "Drag" in this answer actually is more properly described as the "Excess Drag" vector, i.e. Drag minus Thrust)

Using the formula Load Factor = (cosine(arctan(D / (L * cosine bank))) / cosine bank, as derived above, we find the following results:

Bank Angle
L/D
Angle of Descent
Load Factor

0, 0, 0, 0
Inf, 8/1, 4/1, 2/1
0, 7.1, 14.0, 26.6
1, 0.99, 0.97, 0.89

30, 30, 30, 30
Inf, 8/1, 4/1, 2/1
0, 8.2, 12.2, 30.0
1.16, 1.14, 1.13, 1.00

45, 45, 45, 45
Inf, 8/1, 4/1, 2/1
0, 10.0, 19.5, 34.9
1.41, 1.39, 1.33, 1.16

60, 60, 60, 60
Inf, 8/1, 4/1, 2/1
0, 14.0, 26.6, 45.0
2.00, 1.94, 1.79, 1.41

80, 80, 80, 60
Inf, 8/1, 4/1, 2/1
0, 35.7, 55.2, 70.8
5.76, 4.67, 3.29, 1.89

85, 85, 85, 85
Inf, 8/1, 4/1, 2/1
0, 55.1, 70.8, 80.1
11.5, 6.56, 3.78, 1.97

90, 90, 90, 90
Inf, 8/1, 4/1, 2/1
Undefined, 90, 90, 90
Undefined, Inf, Inf, Inf

What can we make of these results?
For mild degradations in the L/D ratio (or more strictly speaking the L / (D -T) ratio), we don't see much decrease in the load factor associated with any given bank angle until the bank angle gets extremely steep, and we certainly don't keep the Load Factor anywhere near 1.  If we degrade the L / D ratio (or more strictly speaking the L / ( D - T) ratio) from infinity all the way down to 4/1-- which is a rather extreme degradation in this ratio-- we still don't see much decrease in the Load Factor associated with any given bank angle until we look at bank angles steeper than 60 degrees.  If we degrade the L / D ratio (or more strictly speaking the L / ( D - T) ratio) from infinity all the way down to 2/1, then yes, so much of the aircraft weight is supported by the Drag vector rather than the Lift vector that we can fly at an 85 degree bank angle with only a Load Factor of 1.97, which is a pretty nifty trick.  But in actual practice, to achieve, such a lousy L/D ratio, we'd probably have to deploy a gigantic drogue chute behind the aircraft, so in large part we're just coming down "under canopy", conceptually speaking!
Looking over the numbers in the table, the idea that we can keep the Load Factor in a very steep bank low simply by cutting power and "dirtying up" the airplane (deploying flaps etc) to degrade the L / (D-T) ratio and increase the descent angle is clearly a fantasy in real-world situations.
And as for the idea that rather than "dirtying up" the airplane, all we need to do is "ease off" the back pressure on the stick-- that idea makes plenty of sense in a dynamic situation like a steep wingover-- nearly all of us have experienced the pleasure of seeing the nose slice down through the horizon with aircraft in a perfect 90-degee bank, while experiencing only a mild G-loading (Load Factor).  But in this situation the flight path is continually curving to point more steeply downward, as the vertical speed rises-- the aircraft is in no way in any kind of steady-state condition.  The dynamics of aerobatic flight are very different from the dynamics of any steady state turn, no matter how steep the bank angle or the descent path and no matter how high the sink rate, and so have little bearing on the original question that should be the focus of the discussion here-- which appears to be whether for any given bank angle, the Load Factor is significantly reduced if the aircraft is allowed to descend at some (possibly quite high) constant rate, allowing some of the weight to be supported by the Drag vector instead of the wing's Lift vector, rather than being kept at a constant altitude.
The idea of easing the G-loading by "easing the back pressure" on the stick deserves a little more attention.  In the short run, a strong decrease in back pressure on the stick will always ease the G-loading, but the decrease in the wing's lift coefficient and Lift force will then make the flight path curve downward and the airspeed increase, so that the wing is once again making about as much Lift (and therefore Load Factor) as it was before the change.  The only long term, sustained effect on Load Factor of a change in amount the back pressure is the pilot exerts on the control stick, lies in resulting change in the L/D ratio.  The table above suggests that that will generally allow only an extremely modest long-term change in Load Factor, for flight at any given bank angle.
And finally, to try to address at least one specific question asked in the original post:

For the setup: In a level turn, 30 degrees would equate to 1.15G and
increase the stall speed by 7%. What are these values if at the
initiation of the turn, the aircraft was in a stable (call it 600 fpm)
descent, and the airspeed was kept constant (relaxing back pressure to
not decrease or increase the speed)?

You mention that you were flying a "deadstick" approach, which suggests you were flying a single-engine light airplane.  Let's assume your approach was made at 70 mph.  If you allowed the plane to descend at 600 fpm, that works out to a descent ratio of 10.27, which gives a descent angle of 5.56 degrees.  Keeping that descent angle firmly in mind, look at the second line of the table above.  Your decrease in Load Factor, compared to the constant-altitude case, is going to be on the order of 1%.  You won't be able to detect any decrease in stall speed due to the fact that you are descending rather than maintaining altitude.  The effect you are asking about is completely insignificant.
And please keep clear that we're not saying that if you feel the stall buffet, you should not ease off on any back pressure you are holding on the stick.  Easing off the back pressure addresses the immediate problem of excessive angle-of-attack, but in the long run cannot change the L/D ratio enough to increase the steady-state descent angle enough to create any significant sustained change the steady-state Load Factor that you'll be experiencing.  Keeping the stick in the new, further forward position will certainly help you avoid stalling, but that is due to the simple fact that you are flying at lower angle-of-attack (which also means a higher airspeed)-- not due to any long-term, sustained decrease in the Load Factor.

I think it finally clicked, thank you! My presumption thus far was that most of the other replies were technically correct but merely keeping to look at graphs didn't change my understanding of them. This does, and is much appreciated! I'll leave it for a short while but am very much inclined to accept it as the answer to my question. Thank you!

Re "But in actual practice, to achieve, such a lousy L/D ratio, we'd probably have to deploy a gigantic drogue chute behind the aircraft, so in large part we're just coming down "under canopy", conceptually speaking!" -- I guess for some reason I was imagining holding the a-o-a and lift coefficient constant at this particular point in the thought experiment, and degrading the L/D ratio only by increasing the drag coefficient. Obviously an aerobatic plane flying straight down at term. vel. (with wings at 0-lift a-o-a) is accomplishing an even bigger decrease in L/D, but in a different way!

The point being that some parts of this answer may not be phrased quite right to be fully accurate in the context of extremely low angles-of-attack and lift coefficients, and near-vertical flight paths, when the wing truly is greatly unloaded by virtue of the forward position of the control stick or yoke.  Will give it some time for ideas to "settle" a bit more before diving in with any changes--

Another thought on the idea of decreasing the G-load (Load Factor) by relaxing the aft pressure on the stick, and my point that (for non-extreme descent angles) most of this decrease can only be temporary and goes away when the airspeed rises-- the reverse thought experiment is to think of starting in a constant -altitude turn and then increasing the G-loading by increasing the back pressure on the stick, while holding the bank angle constant. (ctd)

(ctd) If you are in the part of the flight envelope where increasing your A-o-A also increases your L/D, then this increase in G-load can only be transitory, because the aircraft will transition to a climb, and the G-load in a steady-state climbing turn must be less than the G-load in a steady-state const-alt turn at the same bank angle. In most cases the final decrease in G-load would be due to a decrease in airspeed, compared to the case of the constant-altitude turn at the same angle-of-attack and bank angle.  Eventually this point should be integrated into this answer-- not today though.

Thank you again; although it didn't receive the most votes, it's (to me) the clearest description and the most easily understood, I'll accept it as the answer.

@ExternaUse -- you are welcome!

There's an obvious typo in the first column of the table, 5th row.  I'll wait to change it till making some other planned edits--

"Also, in a steady-state descent, Nclift / Weight = cosine (arctan (D / Nclift)), so Nclift = cosine (arctan (D / Nclift)) * Weight"  -- this arises directly from the closed vector triangle between Nclift, Drag, and Weight -- should point this out explicitly--

If you look at the 4/1 values for "Angle of Descent" in the first three rows of the table, it's clear that there's an error in the first or second row.  Must have been a typo; an actual math error would show up in the "Load Factor" column as well. Planning to run through the math again and edit to correct in the next few days.

Possibly lost in the verbiage of this answer is that for any given bank angle, the G-load will scale according to the cosine of the descent angle.  That point plays a key role in this related answer to a related question -- https://aviation.stackexchange.com/a/8074/34686

@quietflyer  I like your instructive hammer-and-tongs approach.

Due to the descent the load factor wouldn't increase

I have to disagree with your first CFI, even though I suspect that he was misunderstood. Your load factor due to descent will decrease with the cosine of your flight path angle. If you were flying at 60 knots and descending at 600 fpm (which is 30.867 m/s and 3.05 m/s, respectively), your flight path angle was -5.67°, the cosine of which is 0.995. This means your load factor was reduced by 0.5%. This is so insignificant that you might as well neglect it.
Of course, in a vertical dive (cosine of 90° = 0), you can roll any way you  want without having any load factor in the plane-fixed vertical direction at all.

I know you know this, but ... a Minor, I repeat, minor, quibble. The load factor required to maintain your aircraft pitch and bank angle is dependent on the cosine of your flight path angle (as well as your bank angle). But the actual load factor is dependent only on your AOA and Airspeed. (some inertial effects due to changes in flight path angle can also affect it.)

Thanks to you, too, I shall digest this and hopefully ask decent follow up questions.

@CharlesBretana Absolutely! But then you need to maneuver, like pitching up, to add more load factor. I tried to answer the question which is about a descending turn and the load factor change due to the descent.

@Peter, well, just changing pitch attitude, or bank angle, will not change your load factor. The aircraft trim sets the AOA, and the AOA, in combination with airspeed, determines where you are on the Vn diagram, which sets Lift & load factor. So, unless you change the trim, or move the flight control stick (yoke), the load factor will remain, (or change), to whatever aircraft AOA and velocity dictates.

The first CFI is technically correct; you can safely use steep bank angles to descend quickly, taking care not to load the wings any more than required to sustain the turn, but using steep banks to recover a bad base leg setup is not a good idea for a new pilot.
Better to use a slipping turn, or just fly the question mark pattern you used, being careful to keep the turns coordinated and not succumbing to the urge to point the nose at the runway with rudder (a practice that has killed many pilots). If you have passengers on board, you aren't going to be using steep turns or slipping turns, unless you know your passenger is ok with them (they can terrify passengers).
When I tow gliders (Piper Pawnee), I use extremely steep bank angles for the turns to base and final, mostly to maximize my visibility to gliders in the circuit at or near my level, by presenting the wings in planform to them, and I'll roll to near 90 deg and just let the thing fall, pulling just enough pitch to sustain the turn rate. I won't even be at 1G in that case, and with the wings largely unloaded there's little risk of stalling.
But think of it as an advanced maneuver to do when you have experience, and not something you do with passengers.

Not sure if 300 hrs counts as new, but ok.  I didn't necessarily "recover a bad base leg" but used geometry before employing additional tools such as flaps or slips. I don't understand your second para, in how you would propose to fly a slipping turn whilst carefully keeping the turns coordinated. Could you elaborate?
I'd still be interested in the technical background of the manoeuvres that I and you in a way described, and how that lines up with the previous two responses.

Most ppl do question mark turns to final because the were late with the turn initiation. That's what I was referring to. What you were doing was S turning, only once, which is really a normal maneuver on a forced approach.  A slipping turn is definitely uncoordinated, just another version of side slip, but you are keeping AOA low while doing it to reduce stall risk. If you fly an Ercoupe without rudder pedals, S turning to lose alt is the only option on any approach.

Thanks for the clarification; slipping and coordinated didn't quite make sense in your post.

I'm having trouble with this answer.  You state "by presenting the wings in planform to them, and I'll roll to near 90 deg and just let the thing fall, pulling just enough pitch to sustain the turn rate." -- could it be the case that you are not pulling back on the stick enough to hold the glide path angle constant?  Exactly as is arguably the case as you reach a 90-degree bank angle at the apex of a wingover, as is the nose is dropping through the horizon?  That's a totally different situation than a steady-state maneuver.  The G-loading can be (temporarily) much lower in the non-steady-

(ctd) -state case, for any given (instantaneous) combination of pitch attitude and bank angle.  I think the upvoted answer is right in implying that for "normal" aircraft with "typical" drag profiles-- including a glider tug with flaps down and idle power-- the amount of decrease in load factor you can get in a steady-state descent at any given bank angle  by steepening the glide path (by making the drag vector rather than the lift vector support some of the aircraft weight, implying a reduction in airspeed, angle-of-attack, or both) is basically negligible.

When you are banked nearly 90 deg pretty much all the turn rate is coming via the pitch rate, and you are nearly ballistic. You only need to maintain enough elevator to achieve the pitch rate, not support the weight of the machine, and even a pitch rate to turn a very sharp corner is less than that required to hold the airplane up AND maintain the pitch rate.  If you want to turn a sharp corner AND lose a lot of altitude quickly, you can do both by unloading the wings from their "hold the plane up" job by rolling close to knife edge and bringing the nose around with pitch.

A small amount of top rudder can be applied to keep the nose from falling through too fast. It's more or less an extreme slipping turn.

@JohnK -- ok, thanks for the notes, but taking the question into the realm of a slipping turn totally changes the dynamics, as compared to keeping the turn coordinated. Arguably the original question should have been clarified to specify that the turn was coordinated before answers were posted, but s/he is talking about relaxing the back pressure which has essentially no effect on coordination, so s/he doesn't seem to be imagining that the airplane is being made to sideslip.

Well we are really talking about banking steeply to lose altitude. When you roll really steeply like that, you aren't making any effort to sustain altitude with pitch. If you do nothing with rudder, the plane will arc toward the ground as  sideslip from banking keeps the nose aligned to the airflow.

Have some more thoughts re your last-- but right now I'm sitting here trying to remember if there's a way to copy a conversation to chat before the mods move it there themselves...

Guess I'll just let the conversation go on till I get the automated option to move. But as long as you aren't actually moving stick forward to decrease the a-o-a below what you had when flying level, won't the airspeed build up and increase the Load Factor that way?  Sure it may take a few seconds but...  Trying to think of some test I could do soon in actual flight, may be up in a 1-26 soon.  If I slowly roll toward a steeper bank angle, leaving stick at trim pitch-wise rather than pulling back pressure to try to hold altitude, are you saying I won't experience an increase in Load Factor?

As you go steeper and steeper, more of the lift vector is oriented laterally. At 90 deg, there is no vertical component at all, and the airplane becomes a ballistic object, unless you force the fuselage to operate at an angle of attack and make lift, using top rudder.

The load factor is not really the issue here. It is strictly determined by Angle of Attack (AOA) and Airspeed, (i.e., where you are on the Vn diagram.  What is important is AOA. If you lower the nose, (pitch attitude) your vertical velocity will increase, and the loss of potential energy will cause your kinetic energy (airspeed) to increase unless you reduce power significantly. Increasing airspeed allows you to pull more Gs (load factor) at the same AOA, but it does not require that you do so.
If you don't increase your airspeed but want to maintain pitch attitude at a higher bank angle, to generate the higher load factor required for the bank angle, you will need to pull back on the stick or yoke (more back pressure) to increase the AOA to do so.
It is this increased AOA (putting you closer to the stall AOA) that is the problem. The closer to stall (max) AOA you are, the higher of risk of inadvertent stall, and spin (Loss of control) you are.

I'll need  a moment to digest the answers and will report back once I've hopefully understood enough to ask decent enough follow-up questions if there are any.

@quiet flyer, I don't understand your confusion. Increasing airspeed gives you more airflow over the wing. Load factor is Lift / Weight (mass). Lift equation has Velocity squared in it, no? What am I missing in your comment? Are you confused by the word "require"?

If that is your issue, my use of the word "require" is just to make the point that the pilot is controlling the plane's AOA - with the control stick (yoke). Just because the nose goes up or down does not change AOA, (or load factor) all by itself. Indeed, left to itself, as airspeed increases, the AOA will remain more or less constant as determined by aircraft trim, and Load Factor will increase proportionally (as velocity squared).

Maybe the op meant to ask "How does the Load Factor required to maintain aircraft attitude, (pitch & bank), change for an aircraft in a descending turn?"

@quietFlyer... Yes, my Bad, you're correct, I misread your comment. As you stated, "for a given AoA, .... load factor is entirely determined by airspeed.". I deleted that comment...

... but I am still unclear as to what you find to be a problem with "Increasing airspeed allows you to pull more Gs (load factor) at the same AOA, but it does not require that you do so."

@quietFlyer, yes, I see your point now.... As written, the statement is unclear and misleading. To be clear and unambiguous, I should have added why it is not required... to wit, ... "Increasing airspeed allows you to pull more Gs (load factor) at the same AOA, but it does not require that you do so. You could ease off on the back pressure and decrease the AOA "... But I kinda thought that was implied, (and obvious).  Sometimes I assume that readers can take the step in a logical argument without explicitly leading them there.

Actually I guess the root of the problem is that I understood the sentence to consider a hypothetical case where the "givens" were an increased airspeed, and a fixed ("same") angle-of-attack.  But actually you were only thinking of the increased airspeed as the "given".  Now the sentence makes perfect sense to me-- but I wasn't trying to misread it intentionally.  To me something like this would be clearer-- " Increasing airspeed allows you to pull more Gs (load factor) -- this is what will happen if the angle-of-attack is held constant-- but this is not the inevitable result of an incr...

@quietFlyer, Yes... your rewrite also clears up (eliminates) the potential misreading...

This answer considers banking during gliding flight while maintaining a specific rate of descent.   In addition, power needed to maintain the rate of descent is examined.  The airplane evaluated was similar to a 172 having a flight weight of 2550 lb and wing loading of 14.7 lb/sq-ft.  The plane was flown on paper assuming a near-idle, initially neutral, wind-milling propeller at zero thrust. The on-paper conditions easily accommodate powered descent within the constraints imposed. The assisting use of flaps is demonstrated by a low-wing concept aircraft.
The question specifically asks –

What is the load factor in a descending turn?

As posted, the question states two scenarios regarding a descending turn.  The aircraft is described in a power-off, coordinated, descending turn, angle of bank is 30 degrees and gliding at a 600 fpm rate of descent.  Airspeed is kept constant. The question as qualified is very interesting given the constraints regarding angle of bank and rate of descent.  This problem is most easily evaluated by examination of the consequence of forces on the airplane in a banking turn within the gliding-flight definition posed in the question.  In short, what is the aircraft loading factor, n, in a descending, gliding turn, and is that number easily calculated? -

The basic considerations involve forces on the airplane and attitude of the wing. A slightly adjusted sketch by Rina (above) is used to illustrate these forces.  Here, the plane is gliding at lower velocity with approximate 15-degree left bank.  Please note elements depicted in the sketch – the aircraft is not in, but is close to, level-wings gliding flight.  The banked-aircraft flight path is gently curved, but instantaneously, flight is directed along the tangent to the flight-path.  Nevertheless, two aspects of flight happen together for the gliding turn.  First are forces on the aircraft as though incurred in a simple, wings-level glide.  These forces all act through the aircraft center of gravity. They are depicted as $L_g$ and $D_g$ for gliding lift and drag, respectively.  Together, these directed forces are in a vertical plane including the plane’s center of gravity and flight-path tangent.  In sum, these forces, $L_g$ and $D_g$, oppose weight of the aircraft, $W$.  Note that at the center of gravity located within the aircraft, co-ordinate axes x, y, and z intersect.  Stating again, the flight path is directed through the center of gravity, as is spatial orientation of the aircraft.  Note that in simple, wings-level, gliding flight, lift in the vertical plane, $L_g$, is directed forward of vertical by the gliding angle, $\gamma$, of the flight path.
For a simple gliding bank of 15 degrees, little else is changed from a wings-level glide except the orientation of lift, which is now directed off-vertical 15 degrees to the left. The illustration provides an estimated view of flight for our simulated 172 with these conditions. Note the angle of bank, $\phi$, and the lift-oriented vertical plane for a wings-level glide, mentioned above, is also tilted by angle $\phi$, 15 degrees leftward.  This tilted plane also includes the angle of attack, $\alpha$, for the wing, and flight-path tangent.  The flight-path angle, $\gamma$, is still in the real vertical plane, however. The weight of the aircraft, $W$, must now be borne by increased lift $L$ (and drag) incurred in the banked turn. This increase in lift is directly related to the bank angle and is determined in a manner similar to that in level flight. Also note force $F_c$, the centripetal force incurred by the horizontal component of lift, is directing the airplane in a leftward curved path. Lift in the glide, plus centrifugal force (the exact counterpoise of centripetal force in the turn), increase the force of lift incurred in the turn, and consequently increase lift-force loading on the wing.  For conditions shown in the illustration, the proportion by which wing loading will increase, is easily calculated.  This value is given simply  as,
$$n = cos\gamma/cos\phi .$$
However, to calculate this number, given our angle of bank, $\phi$, in descending flight, we must also know our flight-path angle of descent, $\gamma$.   We can estimate this angle simply as the arcsin($v_d$/$V$) were $v_d$ is the vertical velocity on descent, and $V$ is our flight velocity.  For example, descent at 600 fpm is the same as descent at 10 fps, and a flight velocity (air speed) at 66 kts is about 112 fps.  Consequently, the glide slope to be maintained is about 11.2 which is slightly better than the lift-drag ratio of about 10.2 for our airplane in these flight conditions.  This means that power must be added, even though slight, for the aircraft to maintain 600 fpm descent at 66 kts, especially while gliding in a 30 degree bank.  In other words, our airplane has too much drag at 66 kts to match a glide slope of 11.2.  So a bit of power has to be added to flatten our glide and lessen our descent to 600 fpm. The instrument indicated pitch, noted in the analysis, is about +2.3 degrees.
Free-air Flight Performance of the Airplane
Free-air flight considers the airplane flying in a still-air environment.  Most of the aerodynamic calculations assessing flight performance are for the plane in its parcel of air. In the current case, we will not consider wind.  For the airplane under these conditions, the rate of descent in feet-per-minute is easily determined as the flight velocity in feet per second (fps) divided by the lift-drag ratio, $C_L/C_D$, of the airplane, the result of which in fps is multiplied by 60 to obtain rate of descent in fpm.  Consequently, 60sec x 112 fps/10.2, gives the resulting rate of descent, 658 fpm.  In a 30 degree bank at the same flight velocity, if our lift-drag ratio is about 9.3, our resulting rate of descent would be increased to about 720 fpm. The estimated performance of the simulated 172 shows it will not achieve an on-glide rate of descent of 600 fpm across its active range of flight velocities without the addition of power.  Consequently, an attempt to pull up and lessen the rate of descent without the addition of power will exhaust airspeed and result in a stall. This statement is made simply on the basis of the estimated performance of the airplane, and in the interest of safety.  In the abstract, let’s see how this was determined.
First, we should look at the 172 glide performance and the flight-required glide performance across the range of active flight velocities needed to maintain a rate of descent of 600 fpm.  In each of the graphs being shown, the blue line marked by open circles will represent the flight requirement at a rate of descent of 600 fpm, darker blue line with solid circles is the simulated 172 performance in a wings-level descending glide, and red line with open circles is the simulated 172 performance in a descending glide at a 30 degree bank.  An estimated polar for the simulated 172 was used in all of these determinations. The estimated drag was slightly increased in the assessment in order to conservatively portray the required application of additional power (or thrust). -

As shown in the first graph (left, upper), the simulated 172 is able to come close to, but unable to match, the glide ratio needed to maintain a rate of descent at 600 fpm.  The shown minimum rate of descent in a wings-level glide is about 650 fpm at a flight velocity of 66 to 71 kts.  The second graph (left, lower) shows the sustained maximum drag that cannot be exceeded to maintain a rate of descent of 600 fpm.  In a wings-level glide, the simulated 172 clearly exceeds this maximum across its range of performance flight velocities.  The third graph (right) shows the difference between these two measures, namely, the required thrust necessary to flatten the glide slope of the 172 to match a 600 fpm rate of descent.  In other words, in a wings-level glide at its relative minimum gliding velocity of 66 to about 71 kts, the estimated thrust necessary to flatten the glide slope to achieve a 600 fpm rate of descent, is about 16.5 lbs.  This is just a very slight advance on the throttle, just enough (with a little trim) to see the rate of descent decline to 600 fpm.
Parallel conditions for a gliding, 30 degree bank, show the simulated 172 estimated rate of descent is relatively constant at 700 fpm or so in the range of 66 to 89 kts.  Consequently, the necessary thrust to achieve 600 fpm in a 30 degree bank is a modest 68 to 80 lbs.  This simulated 172, by nature of these gentle flying requirements, seems an easy and relatively forgiving airplane to fly.
Do all airplanes fly that way?
We might wonder, are other planes as easy to fly?  Obviously, some are more challenging. Let’s take a look at a low-wing conceptual example somewhat similar to the 172, but being slightly heavier with a wing loading or 16.5 lb/sq-ft. We are, again, considering in parallel, the same aspects for descent at 600 fpm, exactly as done for the 172, but we also consider the use of flaps in a 30 degree gliding turn.  These aspects are indicated just as in the graphic, above, but use of flaps is shown with lighter red features. -

When a similar comparison is made, we can see the concept plane has much lower drag at lower flight velocities and, consequently, a better lift-drag ratio.  These lower-drag factors can present difficulties if trying to bring the aircraft to a 600 fpm rate of descent at flight velocities below about 105 kts.  The plane is simply too aerodynamically clean for the 600 fpm descent glide slope.  Whether in a wings-level glide, or gliding 30 degree bank, the plane can be managed nicely for a 600 fpm descent at airspeeds near 120 kts, or somewhat above.  But at velocities below 105 kts, more drag is needed to manage the glide. Flap deployment will increase drag and should be helpful; let’s see what happens in a 30 degree gliding bank.
For safe deployment in a gliding bank below 90 kts, flaps provide favorable on-glide flight conditions.  The added drag provides needed assistance in managing the glide and rate of descent.  Note that at 66 kts, on-glide conditions at 600 fpm descent are little changed, yet more favorable as power is slightly needed, or not at all, to maintain the rate of descent.  Consequently, at 66 kts the application of slight power is indicated, but below 90 kts, little or no power is indicated. With an adjustable pitch prop, managing drag and maintaining a descent at 600 fpm across this range of flight speeds is not a problem, particularly with the use of flaps below 90 kts.   Above 110 kts flaps are not required, but the gentle application of power is necessary to flatten the in-flight glide slope to maintain a descent of 600 fpm.  Flying this plane without flaps, the pilot must be well ahead of their game.  The use of flaps, however, makes flying this plane much more manageable.

Thanks for the amount of detail. I may be too stupid to understand on how it answers the original question though - quite a few points appear to not have anything to do with the question. I may need somebody to explain the relevance of the information, however well presented it appears at first glance.

@ExternalUse You're a smart fellow! Specifically the load factor in a descending turn is easily calculated. You provided a setup & shown are flight performance characteristics of a Cessna 172 indicating you could not maintain a 600 fpm descent without slight power.  This was simply a general instance along w/comparison to another aircraft.  Aspects shown are for various constant flight speeds. I don’t know what your plane was; this was just an example. The load-factor derivation & glide-slope relationship came from aspects given in the illustration, in case you wanted to investigate further…

@ExternalUse  I definitely appreciate your comments, though.  They are very helpful!

I've merely altered back pressure to keep airspeed stable

This question is interesting as at also touches on steep turn and emergency descent turn techniques.
When one banks 30 degrees, vertical lift is reduced by cosine 30 × G = .866
If airspeed remains the same, angle of attack must be increased by around 15 percent, resulting in a load factor or 1.15 G.
But wait, lift is less than weight in a climb (or descent), as Peter Kampf pointed out. In a shallow descent lift requirement is less than 1 percent less.
So, if one is approaching at 65 knots and raises their stall speed from 50 to 53 knots, there is still a 12 knot margin.
OK, unless a 12 knot tailwind gust comes along.  But you're still better off than the one trying a short field approach at 60 knots.
What is being done here is called an "S" turn, and can indeed be used if the approach is too high.  Excellent that you are coordinated!
But know please in banked turns one can also use increasd speed, rather than increasing AoA, to satisfy the lift requirement.  So, especially low and slow, dip the nose a bit if making a 30 degree turn.  If the plane is staticly stable, it should automatically speed up a little even if you do not re-trim.
So, try it with your CFI.  But you may find the instructor happier if you set up you landing perfectly starting with the downwind, right into a stable, straight final approach.

The 172 answer may be of interest. Also, I have some ideas on resurrecting a closed question, is that ok?

@ThomasPerry  yes, that is fine.

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When pitching, how do planes establish a pitch rate which corresponds to the centripetal acceleration?
Coordinated flight requires that the yaw rate and the centripetal acceleration correspond. In other words, for a given $a_{centripetal} = v^2/r = \omega ^2 r$ this means that the yaw rate must equal $\omega$. We do this by using two independent controls, one for lateral acceleration (linked to the bank angle) and one for yaw rate. When the plane is not coordinated, it is slipping or skidding.
What about in a continuous pitch-wise rotation? In this case, there is only one control, and that's the elevator. In the controls domain this is known as an underactuated system, and it is both theoretically and practically impossible to directly control two system outputs with only one control input.
So if the elevator, by changing the plane's AoA, controls centripetal acceleration, it can't simultaneously control the pitch rate. And the vice-versa is equally problematic, if the elevator controls pitch rate than it cannot directly control centripetal acceleration.
So in a sense, the plane is skidding[*] through the pitch change. So what are the elements which balance the pitch "skid"? The elevator provides a torque, and this torque must 1) ultimately be balanced or else the plane would angularly accelerate and 2) must be counteracted early enough that the plane does not over-rotate.

My hunch is that the wings have such an incredibly strong tendency to align themselves with the airflow that the skidding action is very tiny.
Is this correct?
[*] Skidding, in the sense of a car or ice skater, where the heading vector over-rotates relative to the motion vector. @Koyovis makes the point that for any non-0 AoA, an airplane is always skidding in the vertical direction.

P.S. For a great example of a full-on vertical skid, which occurs when the plane's pitch rate and centripetal acceleration diverge in an extreme manner, check out the SU-27's cobra maneuver

Comments are not for extended discussion; this conversation has been moved to chat.

I may have found something similar, is this kinda what you are asking: https://aviation.stackexchange.com/q/35396/42636

(I've added comments in chat starting here https://chat.stackexchange.com/transcript/message/59229008#59229008 )

Going back to v$^2$/r, think about a centrifuge, or any object orbiting a point.  Is it yawing?  Yaw describes rotation around a specific axis of an aircraft.  Centripetal acceleration is about motion, not orientation.  In any circle, G's are determined by radius and speed, therefor your second "actuator" is throttle or thrust.  In steep turns we go faster and pull harder to get 2G.

Otherwise, the plane simply over-rotates and stalls.  Notice if we go too fast G limits are exceeded.  There's your performance envelope graph.

What about in a continuous pitch-wise rotation? In this case, there is only one control, and that's the elevator. In the controls domain this is known as an underactuated system, and it is both theoretically and practically impossible to stabilize two system outputs with only one control input.

Not if the two outputs are connected. They are not independent of each other!
What connects them is the lift force. All lift that exceeds the weight force is accelerating the airplane into the pitching motion, so that is the centripetal force. Now imagine that this force produces a path that is narrower than what the pitch rate corresponds to: The angle of attack will immediately decline, and so will lift.
Equally, if the pitch rate is higher than the loop produced by the centripetal force, angle of attack will increase, and so will lift.
Again, both are not independent of each other and pitch damping will make sure that both change smoothly through the full trajectory. At least as long as the airplane flies within the limits of linear flight. Once the wing stalls, a different dynamic evolves (where still both are coupled, but now with a hugely varying ratio, depending on angle of attack).

So if the elevator, by changing the plane's AoA, controls centripetal acceleration, it can't simultaneously control the pitch rate.

Yes, it can. AoA determines lift and, therefore, centripetal acceleration. Elevator deflection controls tail lift and, therefore, pitch rate. The elevator does not control AoA directly - it merely trims a pitch rate which moves the AoA to the desired value.
In a first order approximation, all lift is created by the wing and all pitch moments are created by the tail surface. Closer inspection, of course, reveals that the tail also contributes lift and the wing also contributes a pitch moment, but for explaining basic principles let's disregard those niceties. The much larger distance of the tail (or canard) to the center of gravity (CoG) makes its pitch contribution dominant compared to its lift contribution, and vice versa the opposite for the wing.
AoA changes will affect both wing and tail equally, changing both lift and pitch moment. Stability (lift per area) determines whether there is a pitch moment change connected to that lift change.
Pitch changes will also change AoA eventually but add a second component at the tail resulting from pitch damping: Due to its distance from the CoG there is another local AoA and, therefore, lift change on the tail. This lift change is directly proportional to pitch rate and the square of the tail's distance to the CoG, while only the integration of pitch rate over time will change AoA.
Looking at only AoA and pitch rate neglects a third influence: Vertical speed. A change in vertical speed directly changes AoA. Vertical speed changes when there is an imbalance between lift and the sum of weight and centripetal acceleration times mass. The result is a strong damping of any change in AoA which leads to such an imbalance. When you add this effect, you get the coupling between AoA and centripetal acceleration which you seem to have overlooked.

So is the plane skidding through the pitch change?

Depends how you define it. You may view the angle of attack as "skidding", just like a propeller travels less length in one revolution than its blade pitch would suggest. Or you define skidding as a difference between pitch change and change of a tangent to the airplane's trajectory (the first derivative of both over time): Then all skidding is the change in angle of attack due to the change in the direction of gravity (in reference to the airplane) and the change in speed due to the altitude change throughout the trajectory.

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So is the plane skidding[*] through the pitch change?

Yes, absolutely. In fact, if we define skidding in pitch as a difference in pitch angle and heading vector angle, the aeroplane is always skidding when Angle-of-Attack ≄ 0.

My hunch is that the wings have such an incredibly strong tendency to align themselves with the airflow that the skidding action is very tiny.
Is this correct?

No, it is not. Wings don't have a large tendency to align themselves with the airflow, in fact they (and the fuselage) have a strong tendency to position themselves perpendicular to the airflow, that is why an aeroplane needs a horizontal tail.

So if the elevator, by changing the plane's AoA, controls centripetal acceleration, it can't simultaneously control the pitch rate. And the vice-versa is equally problematic, if the elevator controls pitch rate than it cannot directly control centripetal acceleration.

The controlled variable in loops is not pitch rate but the trajectory: a perfect loop is perfectly round, like in OPs picture. This circle trajectory output is achieved using a combination of inputs: airspeed, elevator deflection, load factor (g-forces). As described in this link:

There are loops and then there are round loops. To make a loop round and, well, loop-like, it must be flown with a linear flow and control feel. After you pull back on the stick to enter the loop, you can’t continue to pull the same amount of G all the way around because the speed is decreasing. If you pull 5 G’s with an entry speed of 160 knots, you won’t be able to pull 5 G’s at the apex with a speed of only 100 knots. The pilot has to relax the stick pressure before the top of the loop; then, as the airplane starts downhill again and the speed is increasing, the pilot needs to increase the back pressure so not to over-speed the airplane and to keep the loop round.

+1 things in general don't align themselves with viscous flows very well, if we want that we either need very specific forms or the help of appendages as stated here. For reference: https://www.youtube.com/watch?v=6CtupxjQJCo

@Koyovis, please note that I swapped out the image of the loop with an image of a sinusoidal trajectory. I hadn't asked about a loop, so I think this change of image is consistent with my original question. Unfortunately, since part of this answer focuses on explaining a loop, it winds up looking a little hanging. Sorry for the confusion!

Thanks a lot for the answer. I'm puzzling over the comment about the wings not having a tendency to align themselves. I agree that in isolation the wing will turn perpendicularly, but I think it would be a mistake not to consider the entire airframe. As the wing's aerodynamic center is behind the entire CG, this means that increased lift from an AoA change produces a torque which aligns the airframe with the airflow. Wouldn't that be the stabilizing factor?

Bravo, perhaps the best model is in a non-gravity environment such as the ISS.  Make 2 planes, one of lead (high wing loading) and one of balsa.  The second key is angle of bank.  After a few experiments one will see the direction of relative wind after the deflection from original course determines "skid" or "slip".  In level flight we "skid" through the air to counteract gravity.  Without gravity, it is a pure study of rotation vs sideforce (centripetal force) created by rotation.  Now go back to your horizontal turn, and know why that banked wing can make a slipping turn.

In a loop, the plane will go through varying degrees of skid, coordinated, and slip because of the changing gravity vector.  Level flight is definitely a skid.  Banked turns are... depending on bank, wing loading, adverse yaw, dihedral, tail volume, engine torque ... (just watch that ball and use rudder as needed).

"The increase lift from an AoA change produces a torque which aligns the air frame to the airflow", this is what Peter K is talking about, but alas, the relative wind keeps changing too!  So we have an empirical definition of "coordination": rotation must match rate of turn caused by centripetal force.  In layman's terms, keep the nose into the wind (for lowest drag).  We can see that in pitch, because of gravity, we can't fly along at 0 degrees AoA.

How do aircraft establish a pitch rate which corresponds to the centripetal acceleration.

It is a combination of throttle and elevator, with gravity and aerodynamic forces considered.  If your loop is flown at constant speed, AoA would have to be changed as well as throttle because of the changing position of the gravity vector.
Mathematically, we have the answer in:  G force = v$^2$/r
We can see that airspeed squared and radius are the determining factors.
Starting with the easiest case, straight level flight is where the AoA is enough to cancel gravity (1G, no centripetal force).
Amazingly, it is the banked circle we most commonly fly that may be confusing (banking and yawing), second only to the loop (variable gravity vector), so let's consider the 2 simplest turning models, the pure "skid" turn (0 degrees bank) and the pure "slip" turn (90 degree "knife edge" bank). The mechanisms of skidding and slipping become apparent.
In the case of the horizontal circle, at constant speed with constant throttle, we have that which rotates and that which creates centripetal acceleration (in the direction of the turn).
First, the skidding turn.  With 0 degrees of bank, all we have to accelerate sideways is the crude airfoil created by the fuselage being pitched (yawed) into the turn with the rudder (plus the inward component of the thrust vector).  The rudder is rotating the aircraft to a certain "pitch" to create a certain amount of sideforce and directional stability holds that pitch as the relative wind changes in the turn.  It is the sideways motion created by the airfoil that swings the tail 'round by changing the relative wind.

we can see the relationship between the controlling surface and the stabilizing surfaces is no different in a turn than in straight flight!

On to the pure "slip" knife edge 90 degree bank circle. This time the crude fuselage airfoil (and a lot of power, helps to be light) is holding the aircraft up and the wing is creating lateral acceleration.  Little doubt why many aerobatic pilots like this manuever, by adding in a little pitch with the elevator the wing would rapidly pull it sideways, and again, directional stability swings the tail around, maintaining the pitch set by the elevator.
The common banked turn is perhaps intuitively easier to explain, but we must realize the rudder is simply fine tuning (compensating for adverse yaw) what a good tail should already be doing: following the relative wind.

if the elevator controls pitch (AoA) than it cannot directly control centripetal force

And neither does the rudder in a horizontal turn!.  "Coordinating" a turn merely means reducing drag as much as possible by aligning the aircraft into the relative wind.  The flight path (around the circle) is determined by the thrust and the lateral force vector created by the wing.

"Any more throttle will generate a g force from curving flight upwards." This is incorrect. A force which points normal to the plane's axis of movement will simply accelerate the plane in that direction. The reason the plane follows an arced path is because the force vector is changing in direction, as a result of the plane rotating.

@KennSebesta if you would kindly look at the lift equation, increase in throttle (airspeed) will increase lift.  So will increasing AOA.  So turning your loop is a combination of the two.  Draw in the gravity vectors for entry, top, descent, and bottom.  Since you are actually flying these, I hope you find the right combination that works for you.

The comments on sideslips or skids in the question tend to obscure the fact that no matter what we want to say about yaw rate, turn rate, centripetal force, bank angle, etc, the fundamental cause of a slip or skid is generally that some aerodynamic asymmetry is present which is causing the fuselage to fly in some attitude other than pointing head-on into the airflow.1,2  Thus the "sideways angle-of-attack" of the fuselage is non-zero.  The displacement of the slip-skid ball is the result of this, not the cause.  If we fix this -- often with the rudder-- we'll end the slip or skid.3
And there's simply nothing equivalent to a slip or a skid in regards to purely pitchwise maneuvering, unless we want take the position that any non-zero angle-of-attack of the wing is somehow the "vertical equivalent" of a slip or skid.  Or unless we simply mean flying with the wing at some angle-of-attack that is not the one that yields the flight path we want.  We can fix that by moving the elevator to change the angle-of-attack of the wing  -- problem solved -- assuming that the goal lies within the aircraft's attainable performance envelope!
To a first approximation, the pitch rate we'll see at any point in a maneuver is pretty much "baked in" to the trajectory we are flying, which is determined largely by the combined effects of angle-of-attack, airspeed, and bank angle.4
Perhaps the simplest answer to the question is as follows: "If the intent is to suggest that we can't use the elevator to independently vary the centripetal acceleration and the pitch rotation rate, that is absolutely correct.  Fortunately, we have no need to do this."
Here's another simple answer: "Apply forward or aft pressure to control stick or yoke as needed to manage the angle-of-attack as needed to obtain the desired G-load or the desired pitch rate.  All the other relevant variables will automatically 'come along for the ride'."
Of all the things a pilot might care to concern himself or herself with during aerobatic maneuvering, having "the wrong pitch rate for the centripetal acceleration", or vice versa, really should not be one of them, since we generally can't change one of these parameters without also changing the other.  Of more interest would be g-load versus airspeed, or pitch rate versus airspeed-- and I've actually heard arguments about which of these a pilot should be most concerned with in certain situations such as looping-- (e.g. a hang glider pilot flying loops without the benefit of a g-meter) -- but it's a somewhat academic argument.
Footnotes --

Note that this is possible even if the yaw rotation rate is "correctly" matched to the rate of curvature of the flight path in the yaw dimension, so that the sideslip angle (as indicated by a yaw string) is constant rather than increasing.

Admittedly it's possible to also imagine a situation where a slip or skid is caused not by an aerodynamic asymmetry, but rather by a sudden change in the rate of curvature in the flight path in the yaw dimension, without an adequate yaw torque being applied to change the aircraft's yaw rotation rate to match.  The most familiar example of this would be the pronounced slipping tendency we often see while going "over the top" of a wingover-like maneuver at a steep bank angle, if we choose not to step on the bottom rudder pedal to yaw the nose downward.  At the very top of the wingover, as the aircraft flies in a semi-ballistic trajectory with the wing partially "unloaded", the rate of earthward curvature of the flight path (due to gravity) reaches a maximum value due to the very low airspeed and thus the resulting low radius of curvature of the flight path.  The low airspeed also minimizes the amount of "weathervane" yaw torque that can be generated by the vertical fin.  Unless the aircraft has zero rotational inertia in the yaw axis, this "peak" in the rate of change of the direction of the flight path in the yaw dimension will tend to drive some noticeable amount of sideslip as the nose "lags" behind the actual trajectory of the aircraft.  The yaw string, if present, will tend to blow toward the high wingtip as the aircraft "floats over the top" of the maneuver, and the resulting sideways airflow will then eventually generate the yaw torque that is needed to yaw the nose earthward at a high enough rate to center the yaw string again, or possibly even to drive a brief "overshoot" into a slight skid.  Similarly, when the pitch rotation rate is changing, pitch rotational inertia may sometimes play a role in placing the wing at an angle-of-attack slightly different from the one we'd otherwise "expect" to correlate to the position of the elevator control at any given instant if the aircraft had zero pitch rotational inertia -- see footnote 4 for more.  This will affect the way an aircraft "feels" -- the way it responds to control inputs -- during aerobatic maneuvering.

Note that in general, during turning flight, sharply moving the control stick aft to temporarily "load up" the wing with "extra" g's, or sharply moving the control stick forward to temporarily "unload" the wing to a lower-than-normal g-loading in relation to the bank angle, will immediately curve the flight path up or down as well as decrease or increase the turn radius and rate, but will not cause a significant, immediate change in the position of the slip-skid ball and/or yaw string.  Evidently these control inputs do not create a significant, immediate "demand" for a change in yaw rotation rate -- i.e. an immediate, significant change in the rate of curvature of the flight path in the yaw dimension --so yaw rotational inertia does not tend to drive a significant sideslip.  (Even though, if the new position of the control stick is maintained indefinitely,  the aircraft must eventually settle into equilibrium at a different airspeed, and thus a slightly different turn rate and yaw rotation rate, than it had originally, so that a reduction in angle-of-attack might be expected to promote a slight temporary skid, and an increase in angle-of-attack might be expected to promote a slight temporary slip.  These dynamics don't seem to be observable in actual practice.)  Compare  this to the "wingover" case in footnote 2, where the full three-dimensional evolution of the flight path can eventually drive a rather pronounced sideslip soon after the wing is "unloaded" while climbing, due to the eventual "spike" in the curvature in the flight path in the yaw dimension, and likewise the simultaneous "spike" in the yaw rotation rate that would be required to keep the sideslip angle exactly at zero.

To elaborate further, to a first approximation the elevator can be thought of an angle-of-attack control, just as the rudder can be thought of as a sideslip angle control.  However there are several significant complicating factors.  In the case of the elevator, these factors include pitch rotational inertia (only significant when the pitch rotation rate is changing rather than constant), and the effect of aerodynamic damping of pitch rotation, which also can be viewed as a "curvature" of the free-stream relative wind whenever the flight path is curved.  In general, due to the "curving relative wind" or "pitch damping" effect, the elevator must be raised somewhat higher-- and more "pull" force or less "push" force must be applied to the control stick or yoke -- to achieve the same angle-of-attack whenever the flight path is curving in the pitch dimension (in the nose-up direction, as in a normal turn or in an inside loop), compared to when the flight path is linear.  This is one of the reasons that we often perceive a need to "hold the nose up" in a turn, even in gliders where our goal is to maintain the most efficient angle-of-attack, rather than to maintain a constant altitude.

My chat comments start here https://chat.stackexchange.com/transcript/message/59229008#59229008 ; and see especially https://chat.stackexchange.com/transcript/message/59229754#59229754 and https://chat.stackexchange.com/transcript/message/59229769#59229769 and following posts.

Many of those comments could stand alone as seperate answers to seperate questions- there's really many many questions all rolled into one here. I hope this answer addresses the most important misconception that seems to be at the heart of the original question.

(delete "soon" in 3, change "comments on" to "content regarding")---

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What are the formulas to collimate an image to infinity, on a computer screen?
In a Head Up Display (HUD) or WW2-era "reflector" gunsight, the image is formed at infinity, so you don't refocus your eyes between the HUD, or the sighting reticule, and the outside.
What are the formulas to simulate the same effect on a computer or a smartphone screen ?

What does this have to do with Aviation?


text:
How does vertical stabilisation work?
I know that similar questions are available all over the internet, including this website. However, none of the answers make full sense to me.
I'm looking for simple answer (explain-like-I-am-5 type of answer). The vertical stabiliser has rudder. It provides yaw capablity to the aircraft and I understand that completely. However, people say that if a vertical stabiliser doesn't exist, it's very hard to control the airplane, it rolls the plane out of control if not monitored.
As far as I know, the roll can be controlled using ailerons. I've heard that all wing plane design is not suggested because of lack of vertical stabiliser(s). What exactly is it that the vertical stabiliser does?

Hi sonofel.  Please avoid asking multiple questions in one post.  You should separate out your question about ailerons being used to control roll.

Sure @RyanMortensen, I've edited the question to focus on just one thing.

you should explicit what you don't understand in several similar questions so that we can understand what you don't understand and avoid writing a whole book about aerodynamics.

I suggest you read the part about yaw stability of how it flies and come back with details of what you don't understand.

I read the question one more time; I now see you are saying you understand that the rudder yaws the aircraft.  You aren't saying that you understand yaw stability.  So, maybe my answer is somewhat relevant after all.  Again, I have to emphasize that any discussion of the effect of the vertical fin on roll stability, which seems to be invited by your question, unfortunately has to be above the level of "explain it like I am 5"!

The "vertical stabilizer" (or "vertical fin") provides what is known as "weathervane stability", "directional stability", or "yaw stability".  It makes the plane act like a weathervane.  A weathervane always points into the wind.
The wind a plane "feels" is not the external, meteorological wind, but rather the "wind" created by the plane's movement through the airmass (or perhaps we might prefer to say "within" the airmass).  This apparent wind is called the "relative wind".  So the vertical fin keeps the nose of the plane pointing in the direction that the plane is actually going, i.e. in the direction that the plane is moving through the air.
Imagine driving a car at high speed on an icy lake on a day with no wind.  Can you see how a big vertical fin mounted at the back of the car would provide a "weathervane effect" that would tend to keep the car lined up with the direction it is actually moving, and would tend to prevent the car from "swapping ends" or sliding sideways?
That is what the vertical fin on an airplane does.  It's a very bad thing if an aircraft "swaps ends" or slides sideways through the air at a very large sideslip angle.  In extreme cases, there's essentially nothing the pilot can do to salvage the situation, and the plane may tumble violently out of control.  It's a misconception to say that the function of the vertical fin or vertical stabilizer could be replaced by appropriate control inputs with the ailerons.  It is true in many cases that planes with relatively small vertical fins often require the pilot to make larger rudder inputs to keep turns "coordinated", i.e. to keep the nose from swinging somewhat out of line with the direction the plane is going as it turns -- we'll save saying any more about that for some other answer.
Airplanes that lack a vertical fin must derive "weathervane stability" or "yaw stability" or "directional stability" from other aspects of the aircraft's configuration, generally having to do with the fact that more surface area is behind the center of gravity than in front of it.  For example, a swept-wing configuration generates some amount of "weathervane stability" even without a vertical fin.  Imagine putting a pair of swept wings or delta wings on a pole like a weathervane.  Can you see how, as long as the pivot point were sufficiently forward, the whole assembly would tend to point into the wind like a weathervane?  That's why hang gliders don't need vertical fins.
The question appears to contain a misconception about the relationship between yaw stability and roll stability. In fact, making the vertical fin too large actually tends to make a plane LESS roll-stable or MORE spirally unstable, so that it tends to roll away from wings-level into a steeper and steeper bank and turn.  As a plane starts to roll into a turn, dihedral and sweep (if either are present) tend to generate a stabilizing roll torque that tend to roll the plane back towards wings-level, but only if the developing turn involves some amount of sideslip.  In other words the roll-stabilizing effect of sweep or dihedral are only present if the plane is sideslipping to at least some small degree.  A large vertical fin tends to prevent sideslip, and thus makes the plane behave as if it had less dihedral or sweep. That's why planes that need lots of roll stability, like "free-flight" model airplanes that must fly with no guidance of any kind from either a pilot or a computer, rarely have large vertical fins.  Even though we don't want the plane to completely "swap ends" and tumble, or to slide sideways through the air at a really extreme sideslip angle, we do need for there to be some amount of sideslip in an uncommanded bank and turn in order for sweep or dihedral to be able to bring the plane back toward wings-level.  Likewise when a free-flight model airplane is trimmed for an intentional turn, if the sideslip were somehow completely eliminated, we'd see the bank angle start to get steeper and steeper.  If a pilot were actively controlling the airplane, he could deal with this by making an appropriate control input with the ailerons, but that option doesn't exist with the free-flight model airplane.
So other than preventing a violent tumble when the plane "swaps ends" completely, a vertical fin generally doesn't really contribute to roll stability.

Thank you for such a detailed answer!

The vertical stabilizer is designed to provide a specific amount of passive weathervaning tendency.  Not too much, not too little.  The vertical stab's surface area is key to the airplane's behaviour for both lateral stability and spiral tendencies, and is tied into wing dihedral effect.
You want enough fin surface area to make the airplane passively respond reasonably quickly to yaw deviations, but not too much.  Too little fin area, and its weathervane tendency is too weak and the airplane's nose wanders from side to side in turbulence.
If there is too much fin area, it weathervanes too quickly.  If the airplane is put into a bank by a bump, for dihedral effect to work there has to be a bit of sideslip permitted to generate the self-righting force that dihedral is intended to create.  So fin sizing is intended to allow a bit of lag between the sideslip starting and the weathervaning yaw response, so that dihedral effect can work to level the wings.
If the fin is too big, the weathervaning response is nearly instantaneous so that very little side slip is allowed to develop, and dihedral effect is less effective. So what happens is the airplane is disturbed by a bump and immediately yaws into the low wing and wants to start a descending spiral.
When airplanes are put on floats, the float surface area adds fin surface forward, which has the effect of making the existing fin smaller. The airplane is plenty stable laterally, but the nose wanders in bumps and needs more rudder work to do good coordinated turns (I've flown a Fleet 80 on floats, and it had no extra fin area added to the already smallish vertical fin with floats installed, and if you weren't on top of it with the foot work, it would slither around the sky like a glider would with no rudder inputs).
To fix this, either extra fin area is added at the tail when the floats are installed, or a trick that Cessna uses on later 180s and 185s can be employed, where a bungee spring is added to the rudder, which tends to hold it on center, and makes the rudder contribute some amount to the effective fin area.  You have to live with higher rudder forces however.

Thank you for a detailed answer!

A good way to explain it is to have no wings at all, just use an arrow.  The vertical stabilizer keeps the nose from yawing back and forth just as the horizontal stabilizer keeps the nose from pitching up and down while the arrow is in flight.  This helps it fly straight and true with minimal drag.  Roll is inconsequential until ... we add wings to our arrow.
Now any yaw (from a wind gust) creates a rolling "couple" as one wing goes faster than the other.  Generally not a major issue with straight wings.  Some dihedral takes care of that.
    older than 5 years, please continue reading

Then swept wings and much faster and larger aircraft came along.  Swept wings increase yaw roll coupling by presenting an asymmetric straight half wing and more swept half wing to the airstream in addition to the rolling effect of wing speed asymmetry.
Designers countered by developing very large vertical fins, as seen on the Mig-15.  These presented a whole new host of issues, as now the aircraft was highly asymmetrical to aerodynamic side forces.  Planes of this design tended to spin out of control very easily.
In the rush to the jet age, the vertical symmetry of the biplane, and it's successor, the
high wing monoplane was forgotten.
But, in addition to explaining to the 5 year olds, let them start building and flying for themselves.  They will see how much vertical stabilizer and dihedral they need for their designs.

Shown here:  Jenny's new "friend"

Thank you for such a detailed answer!


text:
How low can a helicopter legally fly over the water with boats below?

My friend was paddling in a big race in Dana Point, California and she said the helicopter was directly over her taking pictures to sell. It looks to be at no more than fifty feet above sea level (if that!); this flight behavior seems insanely dangerous and/or illegal, but maybe I'm wrong. Anyone have hard facts on this type of flying for photos?

Might you wish to insert the word "legally" into the title of your question?

There's no minimum altitude over open water but the pilot must stay at least 500ft away from the boats, per 14 CFR 91.119:

(c) Over other than congested areas. An altitude of 500 feet above
  the surface, except over open water or sparsely populated areas. In
  those cases, the aircraft may not be operated closer than 500 feet to
  any person, vessel, vehicle, or structure.

There's also a catch-all regulation that prohibits endangering other people, 91.13:

(a) Aircraft operations for the purpose of air navigation. No person
  may operate an aircraft in a careless or reckless manner so as to
  endanger the life or property of another.

It's very hard to tell from the photo how far the helicopter is from the boats but if your friend believes it was closer than 500ft, she can report it to the FAA.

I have a feeling there is an exemption in there somewhere for special events though. If you ever watch the chopper that buzzes around F1 races they come real close sometimes.

@Trevor Yes, it's possible the F1 guys have an FAA waiver that gives them an exception under specific conditions.

FAA waivers are common for sporting and "reality" filming.

F1 is an excellent example - was watching the Italian GP a couple of weeks ago and there were shots from a guy in a man-lift looking down on the helicopter. Of course, Monza, Italy isn't covered by FAA regs, but still...

The regulation mentioned above is correct (14 CFR 91.119), but it was taken somewhat out of context. In part D, it says:

(d) Helicopters, powered parachutes, and weight-shift-control
  aircraft. If the operation is conducted without hazard to persons or
  property on the surface—
(1) A helicopter may be operated at less than the minimums prescribed
  in paragraph (b) or (c) of this section, provided each person
  operating the helicopter complies with any routes or altitudes
  specifically prescribed for helicopters by the FAA; and
(2) A powered parachute or weight-shift-control aircraft may be
  operated at less than the minimums prescribed in paragraph (c) of this
  section.

What this means in practice for most of us, is that generally speaking we'll stay at 500-1000, depending on where we're flying, but we often practice auto-rotations and inclined landings in areas off airport, so when taking off and landing, we'll be at much lower altitude. If there's fixed wing traffic, as there often is in Chicago, ATC will direct me to fly at lower than 500 (they'll generally ask me to stay below 500 when I fly along a route also used by fixed wing).
As it was explained to me, we're allowed below as long as we're not conducting ourselves with hazards for the people below, and generally it's more or less using reasoning and common sense. So I can stay relatively low to the ground, but if I started trying to fly between buildings here in Chicago, that would be considered unsafe and a violation. On the other hand, we'll practice off airport landings and maneuvers quite close to the ground, but they'll be in relatively isolated areas (fields, hills, etc.). Over the water, flying at an event like in the picture, if it's condoned/expected for you to be there, it wouldn't be considered a hazard. If you were a private pilot that hadn't requesting it beforehand, I'm guessing you'd be hearing from the FAA.


text:
SVFR required when transitioning through class E to the surface?
So my question is that since we are transitioning through a class E surface area, do we need an SVFR clearance? Part 91.155(c) is not applicable since we are not beneath the ceiling of 1000 ft; however, Part 155(d) states that we cannot enter the traffic pattern of controlled airspace unless we have 3SM visibility or greater. That being said, what dictates how big a traffic pattern is and would ensure we didn't enter it?

My understanding is that it is considered good for "accessibility" (e.g. vision-impaired people using screen readers) if as much as possible of the relevant information is actually included in the text, not only in pictures-- e.g. the printed material at the top of the picture--


text:
Do Airbus avionics give good deadstick landing outcomes?
We have recently had a second good outcome, in Russia, of a no power landing of an Airbus aircraft. The first was the "miracle on the Hudson".
It appears to me that the avionics of this aircraft are very good at handling the stall v. low speed nose up approach problem.
Does anybody have any insight into this programming?

You're referring to this incident, I presume?

"Does anybody have any insight into this programming?" is trivially answered with a "yes" (after all, someone -- likely a number of someones -- wrote that software, hopefully based on a clear specification which itself others would have insight into). Since I presume that isn't the answer you are looking for, what do you really want to know?

The miracle on the Hudson was far from the first dead-stick landing: see https://en.wikipedia.org/wiki/List_of_airline_flights_that_required_gliding

@Eugene Styer, pretty sure he's just asking about Airbus...

Please make some basic research before posting. Basic research make you refer to incident with their flight number, and searching such number into wikipedia gives you insights. Then you may search for report incident on NTSB or BEA websites. This can help you ask clear questions.

@MichaelHall, The list includes all types of aircraft, but there were at last two Airbus incidents prior to the Miracle on the Hudson

@EugeneStyer, it was (as far as that list is correct), however, first such incident of A320-family. The A310 incident is not relevant, because A310 was not fly-by-wire and did not feature the control laws and the while A330, involved in the incident of TS-236, does, I am not sure its flight systems degrade the same way (TS-236 did not have APU as they had no fuel, but I didn't see any mention whether they lost normal law or not).

Airbus (A320+) normal flight law will, at slow speed¹, maintain angle of attack according to the side-stick position, where full aft deflection gets you to the onset of stall.
However the ram air turbine (at least on A320-family; not sure about A330) does not provide enough power to keep all units on and then the system degrades to direct law, in which the side-sticks just control the deflection of the control surfaces directly. This is similar to any other aircraft except since there is no force feedback the side-stick does not become stiffer with increasing speed.
In the US-1549 flight case (landing on Hudson), they had enough time to start the APU, so all systems were powered and they had normal law and did take advantage of it to let the airplane settle as slow as possible.
In the recent U6-178 flight case we don't yet have enough information whether they did have normal law available.

¹ Slower than normal approach speed. At normal speeds, the system maintains wing loading at 1G with neutral side-stick, which can be also described as trimming for current speed. Until the trimmed angle of attack exceeds alpha-limit (12.8° IIRC; 17° is onset of stall, on normal approach should be around half of that), when it switches to maintaining angle of attack.

Note that US1549 did have a bit of a hiccup in that the software would not allow the recommended pitch attitude at full sidestick deflection. This resulted in some unnecessary damage to the aircraft which caused injury to a flight attendant.